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I came across this question and I wasn't sure how to approach it. The question says basically roll a die, what's the probability you get the first 6 on an even throw? I.e., the 2nd, 4th, 6, 8th, etc. die throw. Part of my issue is that obviously you never know when it may come. Here's my guess:

The probability of you getting a 6 on your $n$th throw is:


Now we want to solve when $n=2k$ where $k$ is a natural number which gives us:


The obvious guess is $\frac12$, since it seems it would be equally likely to fall on an even vs. odd, though odd should have an edge since the number of odd rolls are always greater than or equal to the even throws, but unsure.


This point I have no idea how to proceed, have I been doing this right so far? What's the next step?

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The problem with your approach is you're assuming a positive probability on the odd throws to the six. If there is a six on a odd throw it doesn't matter – rlartiga May 4 '14 at 3:37
Do you mean what is the probability that the first 6 comes on an even throw? – soakley May 4 '14 at 3:37
Is that "a 6" or "first 6"? The probability of getting a 6 on the 2nd throw is $1/6$ as there are 6 combinations of 2 throws where the second throw is a 6 out of 36 possible permutations. – JB King May 4 '14 at 3:38
Yes I did first throw, edited to reflect that. – Roger May 4 '14 at 3:45
up vote 7 down vote accepted

Your final expression:

$$a_k = \frac{1}{5} \cdot\left( \frac{25}{36} \right)^k $$

This correctly gives the probability that your first 6 will occur on the $2k$th roll. So, $a_1$ is the probability that your first 6 is on roll 2; $a_2$ is the probability that your first 6 is on roll 4; etc.

In order to find the probability $p$ that your first 6 is on any even roll, we just need to add up all the $a_k$s:

$$p = \sum_{k=1}^\infty a_k = \sum_{k=1}^\infty \left[ \frac{1}{5} \cdot \left( \frac{25}{36} \right)^k \right]$$

The formula for the sum of an infinite geometric series with a ratio $r < 1$ (of the form $a_1, a_1 r^1, a_1 r^2, \dots$) is well-known:

$$\sum_{k=1}^\infty a_1 r^k = \frac{a_1}{1-r}$$

Comparing the form of an infinite geometric series to the formula we've come up with for $a_k$, we see that $a_1 = (1/5)(25/36)$ and the ratio is $r = (25/36)$. Hence:

$$p = \frac{a_1}{1-r} = \frac{25/180}{11/36} = \frac{5}{11}$$

This is close to $1/2$, but not exactly equal to it.

One way to intuitively understand why this is is to think of this as a "game" between two players who take turns rolling the die. In each pair of turns, the first player to roll automatically has an advantage, since if they roll a 6, they win, whereas the second player only gets a chance to roll if the first player hasn't already won. This is why it's less likely that you get the first 6 on an even roll than on an odd roll - the even rolls correspond to the "second player", who is at a disadvantage.

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Thank you so much, this is a great answer. – Roger May 4 '14 at 5:47

It is intuitively clear, and not hard to show, that the game terminates with probability $1$. Let $a$ be the probability that the game ends on an odd throw. Then our desired probability is $1-a$.

On the first toss, with probability $\frac{1}{6}$ the game ends immediately, and hence on an odd throw. With probability $\frac{5}{6}$, the game does not end immediately, and therefore the probability the game ends on an odd throw is $1-a$. Thus $$a=\frac{1}{6}\cdot 1+\frac{5}{6}\cdot(1-a).$$ Solve the linear equation for $a$. We get $a=\frac{6}{11}$, so the probability the game ends on an even throw is $\frac{5}{11}$.

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It is easiest to visualize as a two-player game, A and B tossing alternately, A first. Then odd means first player wins. Suppose A doesn't toss a $6$ on first throw. Then it's a new game, with B going first, so probability A wins is probability that – André Nicolas May 4 '14 at 9:37
"first" player loses. Or to put it in terms of odd and even, if on first toss we don't have a $6$, then it's a new game, except that because of the first toss the roles of odd and even have reversed. – André Nicolas May 4 '14 at 9:46
I see! Thank you for that clarification, for some reason I didn't see it until now. That was a brilliant approach to this problem! – Roger May 4 '14 at 17:39
You are welcome. The idea (conditioning on the first "toss") has a number of uses, particularly in the computation of expectation. – André Nicolas May 4 '14 at 17:42

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